Difference between revisions of "2014 AIME II Problems/Problem 14"
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<math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math> or <math>AP=2HN=HM</math> | <math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math> or <math>AP=2HN=HM</math> | ||
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| + | Now we know that HM=AP, we can find for HM which is simpler to find. | ||
| + | |||
| + | We can use point B to split it up as HM=HB+BM, | ||
| + | |||
| + | We can chase those lengths and we would get | ||
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| + | <math>AB=10</math>, so <math>OB=5</math>, so <math>BC=5\sqrt{2}</math>, so <math>BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}</math> | ||
Revision as of 21:39, 29 March 2014
14. In △ABC, AB=10, ∠A=30∘, and ∠C=45∘. Let H, D, and M be points on the line BC such that AH⊥BC, ∠BAD=∠CAD, and 
. Point 
is the midpoint of the segment 
, and point 
is on ray 
such that PN⊥BC. Then 
, where 
and 
are relatively prime positive integers. Find 
.
http://www.artofproblemsolving.com/Wiki/images/5/59/AOPS_wiki.PNG ( This is the diagram.)
As we can see,
is the midpoint of 
and is the midpoint of
is a 
triangle, so ∠HAB=15∘.
is 
.
and 
are parallel lines so 
is also.
Then if we use those informations we get 
and
and 
or 
Now we know that HM=AP, we can find for HM which is simpler to find.
We can use point B to split it up as HM=HB+BM,
We can chase those lengths and we would get
, so 
, so 
, so 
